# About nondegeneracy of Killing form

Let $$\mathfrak{r}$$ be a reductive Lie subalgebra of the complex semisimple Lie algebra $$\mathfrak{g}$$ and let $$B$$ be the Killing form of $$\mathfrak{g}$$.

Suppose that the restriction $$B|_\mathfrak{r}$$ of B on $$\mathfrak{r}$$ is non-degenerate. Let $$\mathfrak{g} = \mathfrak{r} \oplus\mathfrak{s}$$ be the orthogonal decomposition with respect to $$B$$.

How to check that the restriction $$B|_\mathfrak{s}$$ of $$B$$ on $$\mathfrak{s}$$ is also non-degenerate?

• Hint: the matrix associated to the form is block diagonal when you pick bases from those two subspaces. Feb 26, 2019 at 8:52
• The accepted answer is of course correct, but maybe one should notice that this has very little to do with Lie algebras. It's a general fact that if $B$ is a non-degenerate bilinear form on a vector space $V$, and $W$ is a subspace, then the restriction of $B$ to $W$ is non-degenerate if and only if $V = W \oplus W^\perp$. Cf. math.stackexchange.com/q/1295105/96384 or math.stackexchange.com/q/3550381/96384. Here you just use the $\Leftarrow$ direction (which of course is proved most easily as in Tsemo Aristide's answer). Mar 16, 2022 at 18:48

Suppose $$g$$ is the orthogonal sum $$g=r\oplus s$$ where $$g$$ is semi-simple. Since $$g$$ is sem-simple, Killing is not degenerated, this implies that for every $$y\in s$$, there exists $$z\in g$$ such that $$B(y,z)\neq 0$$, we can write $$z=u+v, u\in r, v\in s$$, we deduce that $$B(y,z)=B(y,u+v)=B(y,v)\neq 0$$ since $$r$$ is orthogonal to $$s$$. This implies that the restriction of $$B$$ to $$s$$ is not degenerated.